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The peripheral system is clearly antiaromatic if planar, but we know that antiaromatic compounds are more unstable than non aromatic compounds.

So, I think that the molecule will lose its planarity and become non aromatic, but this molecule is antiaromatic according to my professor.

Isn't my reasoning correct?

Also, what would be the IUPAC name of this molecule?

  • $\begingroup$ "planar" is relative. this molecule is likely planar enough to give enough pi orbital overlap for antiaromaticity $\endgroup$ – Karl Sep 18 '18 at 19:17
  • $\begingroup$ Yes, this is what my professor further said while explaining. It's not just about the planarity but about the pi conjugation which will be reinforced when the middle carbon lifts out of the plane bringing the pi orbitals closer. $\endgroup$ – ad_infinitum Sep 19 '18 at 13:35
  • $\begingroup$ According to chemdoodle the name is 13-Methyltricyclo[,13]trideca-1(12),2,4,6,8,10-hexaene. $\endgroup$ – matt_black Sep 21 '18 at 15:20
  • $\begingroup$ A little information: check out structure 6 on this Wikipedia page and read the description below it. Also, if you have access, see the paper it references. $\endgroup$ – SendersReagent Oct 18 '18 at 21:10

The central atom is approximated to be tetrahedral and thus approximately $\mathrm{sp^3}$ hybridised. The three atoms of the outer ring it is bonded to are approximately $\mathrm{sp^2}$ hybridised as they are all part of double bonds.

If you build this, you will notice that while these three atoms are necessarily in a plane they have different orientations. It is simply not possible to make a planar ring without some significant distortions of the geometry. I thus challenge your professor’s assumption.

I would expect this system to have strongly localised double bonds, i.e. significant bond length alteration, in such a way that the double bond are twisted with respect to each other to minimise strain.

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